Question

The electric potential due to infinite uniformly charge of linear charge density $\lambda $ at a distance r from is given by is the reference point ${{r}_{o}}$
$\begin{align}
  & a)\dfrac{\text{2 }\!\!\lambda\!\!\text{ }}{\text{4 }\!\!\pi\!\!\text{ }{{\in }_{\text{o}}}r} \\
 & b)\dfrac{\text{2 }\!\!\lambda\!\!\text{ }}{\text{4 }\!\!\pi\!\!\text{ }{{\in }_{\text{o}}}{{r}_{0}}} \\
 & c)\dfrac{\text{2 }\!\!\lambda\!\!\text{ }\left[ \ln {{r}_{o}}-\ln r \right]}{\text{4 }\!\!\pi\!\!\text{ }{{\in }_{\text{o}}}} \\
 & d)\dfrac{\text{2 }\!\!\lambda\!\!\text{ (lnr)}}{\text{4 }\!\!\pi\!\!\text{ }{{\in }_{\text{o}}}r} \\
\end{align}$

Answer 1

Hint: The electric field due to an infinitely charge distribution is given by, $\dfrac{\lambda }{2\pi {{\in }_{\circ }}r}$ where $\lambda $ is the linear charge density, ${{\in }_{\circ }}$ is the permittivity of free space and r is the perpendicular distance from the charge distribution. Hence taking ${{r}_{o}}$ as the reference point, the potential at the point which is at a distance r from the charge distribution from the expression i.e. $\dfrac{dV}{dr}=\dfrac{{{V}_{r}}-{{V}_{{{r}_{o}}}}}{dr}=-E...(1)$ where ${{V}_{r}}$ is the potential at point r, ${{V}_{{{r}_{o}}}}$ is the potential at the reference point and E is the electric field at point r.

Complete step-by-step answer:

In the above figure we can see a charge distribution of linear charge density $\lambda $. Now we wish to calculate the potential at point r from the linear charge distribution. For that let us use equation 1 to determine the potential at point r with respect to the reference point ${{r}_{o}}$. Let us assume the distance between them is very small i.e. dr.
$\begin{align}
  & \dfrac{{{V}_{r}}-{{V}_{{{r}_{o}}}}}{dr}=-E,\text{since }E=\dfrac{\lambda }{2\pi {{\in }_{\circ }}r} \\
 & {{V}_{r}}-{{V}_{{{r}_{o}}}}=-\dfrac{\lambda }{2\pi {{\in }_{\circ }}r}dr \\
\end{align}$
Now let us integrate the above equation to find the potential at point r with respect to ${{r}_{o}}$.
$\begin{align}
  & {{V}_{r}}-{{V}_{{{r}_{o}}}}=-\int\limits_{{{r}_{o}}}^{r}{\dfrac{\lambda }{2\pi {{\in }_{\circ }}r}dr} \\
 & {{V}_{r}}-{{V}_{{{r}_{o}}}}=-\dfrac{\lambda }{2\pi {{\in }_{\circ }}}\int\limits_{{{r}_{o}}}^{r}{\dfrac{1}{_{\circ }r}dr} \\
 & {{V}_{r}}-{{V}_{{{r}_{o}}}}=-\dfrac{\lambda }{2\pi {{\in }_{\circ }}}\left[ \ln r \right]_{{{r}_{o}}}^{r} \\
 & {{V}_{r}}-{{V}_{{{r}_{o}}}}=-\dfrac{\lambda }{2\pi {{\in }_{\circ }}}\left[ \ln r-\ln {{r}_{o}} \right] \\
 & {{V}_{r}}-{{V}_{{{r}_{o}}}}=\dfrac{\lambda }{2\pi {{\in }_{\circ }}}\left[ \ln {{r}_{o}}-\ln r \right] \\
\end{align}$
Hence from the above obtained result we can conclude that the potential at point r with respect to the reference point is given by,$\dfrac{\lambda }{2\pi {{\in }_{\circ }}}\left[ \ln {{r}_{o}}-\ln r \right]$ .

So, the correct answer is “Option C”.

Note: If we consider the above equation the potential keeps on decreasing as we move away from the reference point ${{r}_{o}}$. Hence we always take the electric field as a negative gradient of potential between the two points. The potential at a point always varies inversely with the distance.